I know I must use 3D integrals and Fourier moments but I don't know how I can do that !Please someone give me some clues about a possible math proof ?

Thanks by advance

PS: I already try to find this formula with the gauss theorem as said in the older topic but I never find the result but an approximate result! I obtain the formuma here but with some addition in more...

Thanks to your response
but I don't see how it is a "simple" result of the divergence theorem because you don't compute with element surface but with tetrahedron where the base is each triangle of the parametrized surface and the common point is the origin of the 3d space (0,0,0).

The divergence theorem said that the volume is equal to the outward flux of the surface, isn't it ?
Here we don't use surface but volume of tetrahedron composed with origin ?

Take the vector field: F=(x-x0)/3, x and x0 are in R^3.
Its divergence with respect to x is 1.

So: Volume = Int[Div(F)*dV] = Int[<F,dS>] = Sum[A_i * (d_i - <n_i,x_0>)]/3
Where: A_i = Area of triangle. With its plane equation <n_i,x>=d_i, with unit normal.
If you choose x0=0, then you get Sum[A_i*d_i/3]
For unit n_i, d_i is exactly the distance of the plane from the origin. So what is written in the sum is exactly the volume of the tetrahedron.

If you take vector field F (x,y,z) -> (x-x0,0,0)/3 the divergence is 1/3, not 1 ??
Moreover I don't understand how you insert your inner product d_i - <n_i,x0> and directly the area of a triangle after the gauss theorem
=> Int[<F,dS>] = Sum[A_i * (d_i - <n_i,x_0>)]/3 ??

x and x0 are points in R^3 and not coordinates.
As for the second question:

F*dS is a two-form defined over the boundary of the volume, while <F,n>*dA is the corresponding two-form defined over subset of R^2.
F*dS = <F,n>*dA = (<x,n> - <x0,n>)/3*dA = (d - <x0,n>)/3*dA
Now, d - <x0,n>/3 is a constant within the triangle, so the integral is: A*(d - <x0,n>)/3.

I've been abusing notation here a bit. It makes little sense to do an inner product between 1-forms and vectors, but it should be fairly obvious what I meant in <~,n>.

I don't know it was possible to use points in the function F ? I believed it was just vector field ?

I have a last question (in my mind easy but I want to be sure) : I use only vertices in the code above (v0 v1 v2).
Or your variable d is a distance not a point.
How can we say that A*(d-<x0,n>)3 == 1/2 (P1^P2).d/3 where d must be a vertice ? (n is not obligatory (1,1,1)?)
Moreover if i use the formule of area it's also distance and not vertices....

Well, vector fields assign vectors to points. So it's only natural to use points in them.
As for 'd', then yes, it's the distance from the plane to the origin when unit normal is used. Since during the integration 'x' lies on the plane, <x,n> is constant and it equals exactly to 'd'.

I haven't been following the thread, but the code posted in the first post seems fine to me. Instead of "Dot(v0,Cross(v1,v2))" I would have written "Determinant(v0,v1,v2)", but other than that it's exactly how I would have written it.

Yes So it doesn't prove my formula with vertices ??The aim of this formula is to not use distance but just vertices to gain time !

Why not?For tetrahedron: A*d/3 == Det(v1-v0,v2-v0,v3-v0)/6, where 'A' is the area of some triangle and 'd' is the distance from the opposing vertex to the plane of the triangle. In the Det formula v1,v2,v3 define the same triangle, while v0 is the opposing vertex.

Yes So it doesn't prove my formula with vertices ??The aim of this formula is to not use distance but just vertices to gain time !

Why not?For tetrahedron: A*d/3 == Det(v1-v0,v2-v0,v3-v0)/6, where 'A' is the area of some triangle and 'd' is the distance from the opposing vertex to the plane of the triangle. In the Det formula v1,v2,v3 define the same triangle, while v0 is the opposing vertex.

One more time thanks !

But ... I have one more question?? (yes again)I agree with you for the Det but in the proof we have already use the origin for x0 ! Or in my formula the origin will be your V0 in the det formula as the opposing vertex for each triangle in the mesh. It's not a wrong thing to use the origin for the X0 in F and for the opposing vertex. In my mind , it's look like a ruse... no ?

I'm not sure whether I got you right, but why is it a bad thing to use the origin? You can choose x0 to be whatever you want, and as long as it's constant this formula will work.

I'll try to rephrase. In the vector field F you can choose x0 as any point in R^3. After some manipulations we can interpret the result as the sum of volumes of multiple tetrahedrons that are comprised of the faces of the mesh and have one vertex in common, that is x0.
Geometrically this makes a lot of sense for sphere-like (or more generally convex) meshes, just choose x0 to be somewhere within the object and the formula should come trivially to you without the divergence theorem. For arbitrary oriented manifolds (or when x0 is not within the enclosed volume) this is not so clear, but the divergence theorem takes care of that.

Yep,
I understand your last topic and that's why I decided to use this theorem (it's a very interesting thing).
But my worry is completely in the proof because we use the origin in the vector field (make a lot of sense to simplify the formula) AND I use the origin for one of my vertice (opposing vertex) in the Det formula to obtain the dot product of a cross product of three points (after simplification)...

Or in the formula we obtain A*d/3 where d == <x,n> because (<x,n>-<x0,n>)/3==<x,n>/3. To simplify the det formula I use x (opposing vertex)=Origin=xo. And it's here my problem....

This is not a problem since this is exactly the same point.Let's recap it by steps:1. You start with the volume integral.2. Afterwards, you choose a vector field whose divergence is 1. This vector field is defined up to a constant x0.3. As any x0 is good, you choose one (for instance the origin) and obtain the flux integral by the divergence theorem.4. Once finished computing the flux integral, you choose to interpret the formula as the sum of volumes of tetrahedrons that share the vertex x0.5. You rewrite the formula from step 4, with an equivalent formulation that uses Det. Again, this is just a change in interpretation.